Answer:
The second system must be set in motion seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion seconds later
Answer:
1.53 m/s
Explanation:
Given:
Mass of the car (M) = 1300 kg
Mass of the coal (m) = 400 kg
Initial velocity of the car (U) = 2 m/s
Initial velocity of the coal (u) = 0 m/s (Since it is dropped)
When the coal is dropped into the car, then they move with same final velocity.
Let the final velocity be 'v' m/s.
For a closed system, the law of conservation of momentum holds true.
So, initial momentum is equal to final momentum of the car-coal system.
Initial momentum of the car =
Initial momentum of the coal =
Total initial momentum is the sum of the above two momentums.
So, total initial momentum = 2600 + 0 = 2600 Ns
Now, final momentum is given as the product of combined mass and final velocity. So,
Final momentum of the system =
Now, from law of conservation of momentum,
Initial momentum = Final momentum
Therefore, the final velocity of either of the two masses is same is equal to 1.53 m/s.